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I am trying to figure out the formula for RSS-ing two parallel resistors, but for the life of me can't figure this out.

For series resistors it is easy, just $${R_1 + R_2} \pm \sqrt{ dR_1^2 + dR_2^2}$$

But what is it for two parallel resistors where you use the formula $$R_{\text{tot}} = {R_1 \cdot R_2\over R_1+R_2}$$?

I was thinking you just work out \$\small R_1 \cdot R_2 \pm \text{tol}_A\$ and \$\small R_1+R_2 \pm \text{tol}_B\$, then just do the division $$R_{\text{tot}} = {R_1 \cdot R_2\over R_1+R_2}$$ and $$\text{tol} = {\text{tol}_A \over \text{tol}_B}$$ but this is not right? Does anyone know how to do this?

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  • \$\begingroup\$ The search term is "propagation of uncertainty or error": geol.lsu.edu/jlorenzo/geophysics/uncertainties/…. Break down each part of the divider equation and apply each rule. You can also just use sums of partial derivatives which is where the rules come from. \$\endgroup\$
    – DKNguyen
    Commented Dec 14, 2021 at 14:49
  • \$\begingroup\$ Hi DKNguyen, this is not RSS though. RSS gives a more realistic estimate for the error. This document is talking about Worst Case Analysis (WCA) \$\endgroup\$
    – Edba
    Commented Dec 14, 2021 at 15:46
  • \$\begingroup\$ @Edba but this is not right - you didn't say what it should be - what should it be? \$\endgroup\$
    – Andy aka
    Commented Dec 14, 2021 at 15:47
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    \$\begingroup\$ Read it more carefully. \$\endgroup\$
    – DKNguyen
    Commented Dec 14, 2021 at 15:48
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    \$\begingroup\$ Convert to conductances and they add. Note that your example values, the 20K adds 25x the uncertainty of the other. In your comment, the 1.414% error is only true for approximately equal resistors; if one R is sufficiently larger (in series) or smaller (in parallel) you can practically ignore the other's tolerance. \$\endgroup\$
    – user16324
    Commented Dec 14, 2021 at 20:45

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Does anyone know how to do this?

If I make a comparison using worst case analysis (with both parallel resistors being very similar in their basic value), it's the same for series resistors too.

So, if you are able to convert a WCA figure to an RSS figure then, the result is the same.

I mention that both resistors should be similar in their basic value because, that will be the most onerous situation to consider.

In series

$$R_{NET} = R_1(1 \pm\Delta)+R_2(1 \pm\Delta)$$ $$= (R_1+R_2)\cdot (1 \pm\Delta)$$

In parallel

$$R_{NET} = \dfrac{R_1(1 \pm\Delta)\cdot R_2(1 \pm\Delta)}{R_1(1 \pm\Delta)+R_2(1 \pm\Delta)}$$ $$= \dfrac{R_1\cdot R_2}{R_1+R_2}\cdot (1 \pm\Delta)$$

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  • \$\begingroup\$ Could anyone tell me why the above has been downvoted. I'd like to fix the answer if there's a mistake but I've probably gone blind to the issue. \$\endgroup\$
    – Andy aka
    Commented Dec 17, 2021 at 5:50

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